# -7.8.1 Amp, Period, PS, VS of 3cos(\pi x-2)+5

• MHB
• karush
In summary, we discussed the plug-in equations used to find the amplitude, period, phase shift, and vertical shift in a graph with the given equation. We also clarified the symbols used and determined that the period is equal to 2 and the phase shift is equal to 2 divided by pi.
karush
Gold Member
MHB
Find amplitude, period, PS, VS. graph 2 periods of
$y=3\cos(\pi x-2)+5$

ok I think these are the plug ins we use
$Y=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B$
or
$A\cos\left(\omega x-\phi\right)+B$
A=amplitude B=VS or veritical shift
$T = \dfrac{2\pi}{\omega-\phi}$
$PS = 0$ assumed here

ok just want to see if I have these plug in eq right, different books use different symbols

karush said:
Find amplitude, period, PS, VS. graph 2 periods of
$y=3\cos(\pi x-2)+5$

ok I think these are the plug ins we use
$Y=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B$
or
$A\cos\left(\omega x-\phi\right)+B$
A=amplitude B=VS or veritical shift
$T = \dfrac{2\pi}{\omega-\phi}$
$PS = 0$ assumed here

ok just want to see if I have these plug in eq right, different books use different symbols
Use $$\displaystyle Y = A\cos\left(\omega x-\phi\right)+B$$ or $$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$$. (You had one too many x's in your first equation.)

-Dan

$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$
then for $y=3\cos(\pi x-2)+5$
$A=3 \quad \omega=\pi \quad \phi=2 \quad B=5$
before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$

karush said:
$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$
then for $y=3\cos(\pi x-2)+5$
$A=3 \quad \omega=\pi \quad \phi=2 \quad B=5$
before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$
Yup. :)

-Dan

before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$
so then
$T=\dfrac{2\pi}{\pi}=2$ and $PS=\dfrac{2}{\pi}$
kinda ? on PS
So T is Period?

karush said:

before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$
so then
$T=\dfrac{2\pi}{\pi}=2$ and $PS=\dfrac{2}{\pi}$
kinda ? on PS
So T is Period?
Yes. You have it right.

-Dan

## 1. What does "-7.8.1 Amp" mean in the equation?

The "-7.8.1 Amp" refers to the amplitude of the cosine function. It represents the maximum displacement from the equilibrium position. In this case, the amplitude is -7.8.1 units.

## 2. What is the period of the function?

The period of the function is the length of one complete cycle of the cosine curve. In this equation, the period is 3 units.

## 3. What do "PS" and "VS" represent in the equation?

"PS" stands for phase shift, which is the horizontal displacement of the cosine curve from its original position. "VS" stands for vertical shift, which is the vertical displacement of the curve. In this equation, the phase shift is -2 units and the vertical shift is 5 units.

## 4. How do I graph this equation?

To graph this equation, plot points on a coordinate plane by substituting different values for x and solving for y. Then, connect the points to create a smooth curve. The amplitude, period, phase shift, and vertical shift can help determine the shape and position of the graph.

## 5. What is the significance of the number \pi in the equation?

The number \pi in the equation represents the ratio of a circle's circumference to its diameter. In this context, it is used to calculate the horizontal displacement of the cosine curve due to the phase shift. It is a constant value of approximately 3.14.

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